Question

if ( f ) is the function defined by ( f(x) = \frac{x^2 - 1}{sqrt{x} - 1} ), then ( lim_{x \to 1} f(x) ) is
a 4
b 2
c 0
d nonexistent
answer c
incorrect. this results if the function ( f ) is incorrectly simplified to ( \frac{x^2}{sqrt{x}} - \frac{1}{1} ). the limit of this expression as ( x \to 1 ) is

Explanation:

Step1: Factor the numerator

We know that \(x^{2}-1\) is a difference of squares, so \(x^{2}-1=(x - 1)(x + 1)\). Also, we can rationalize the denominator or rewrite \(\sqrt{x}-1\) in terms of a difference of squares. Let \(y=\sqrt{x}\), then when \(x = 1\), \(y = 1\), and \(\sqrt{x}-1=y - 1\), and \(x-1=y^{2}-1=(y - 1)(y + 1)\). So we can rewrite the function:
\[
f(x)=\frac{x^{2}-1}{\sqrt{x}-1}=\frac{(x - 1)(x + 1)}{\sqrt{x}-1}=\frac{(y^{2}-1)(x + 1)}{y - 1}=\frac{(y - 1)(y + 1)(x + 1)}{y - 1}
\]
For \(x
eq1\) (since when \(x = 1\), the original denominator is \(0\)), we can cancel out \(y - 1=\sqrt{x}-1\) (as long as \(\sqrt{x}-1
eq0\), i.e., \(x
eq1\)). After cancellation, we have \(f(x)=(y + 1)(x + 1)\) where \(y=\sqrt{x}\), so \(f(x)=(\sqrt{x}+1)(x + 1)\) for \(x
eq1\).

Step2: Evaluate the limit as \(x

ightarrow1\)
Now we find the limit as \(x
ightarrow1\) of \(f(x)=(\sqrt{x}+1)(x + 1)\). We substitute \(x = 1\) into the simplified function (since the limit as \(x
ightarrow1\) does not depend on the value of the function at \(x = 1\), only the values near \(x = 1\)):
\[
\lim_{x
ightarrow1}(\sqrt{x}+1)(x + 1)=(\sqrt{1}+1)(1 + 1)=(1 + 1)\times2=2\times2 = 4
\]

Answer:

A. 4